A light year is a measure of O length O speed O time O volume

Why do rats like cheese? <>-<> Im freaking out here dudes

AVprozaik [17]

They like cheese because it is easy to eat and it tases really good

7 0

2 years ago

Read 2 more answers

What are the units of k in the following rate law?

Ludmilka [50]

Answer:

B. $\frac{1}{M^{2} s }$

Explanation:

The unit for rate is M/s while the unit for each molecule should be M. You can find the unit for k by putting the units for rate and the molecules into the equation

rate= k{X][Y]

M/s= k * $M^{2}$ * $M^{1}$

k= (M/s) / ( $M^{3}$ )

k= $\frac{1}{M^{2} s }$

You can also use this predetermined formula to solve this problem faster: k= $\frac{M^{1-n} }{s }$

Where n is the number of molecule. There are 3 molecule(2X and 1Y) so n=3, so

k= $\frac{M^{1-n} }{s }$

k= $\frac{M^{1-3} }{s }$ = $\frac{m^{-2}}{s}$ = $\frac{1}{M^{2} s }$

8 0

3 years ago

Be sure to answer all parts. Write the equations representing the following processes. In each case, be sure to indicate the phy

barxatty [35]

Answer:

1. $S^{-}(g)+e^{-} \rightarrow S^{2-}(g)$

2. $Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}$

3.The electron affinity of $Mg^{2+}$ is zero.

4. $O^{2-}(g) \rightarrow O^{-}(g)+e^{-}$

Explanation:

1.

Electron affinity:

It is defined as the amount of energy change when an electron is added to atom in the gaseous phase.

The electron affinity of $S^{-}$ is as follows.

$S^{-}(g)+e^{-} \rightarrow S^{2-}(g)$

2.

Ionization energy:

Amount of energy required to removal of an electron from an isolated gaseous atom.

The third ionization energy of Titanium is as follows.

$Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}$

3.

The electronic configuration of Mg: $1s^{2}2s^{2}2p^{6}3s^{2}$

By the removal of two electrons from a magnesium element we get $Mg^{2+}$ ion.

$Mg^{2+}$ has inert gas configuration i.e, $1s^{2}2s^{2}2p^{6}$

Hence, it does not require more electrons to get stability.

Therefore,the electron affinity of $Mg^{2+}$ is zero.

4.

The ionization energy of $O^{2-}$ is follows.

$O^{2-}(g) \rightarrow O^{-}(g)+e^{-}$

3 0

2 years ago

How many grams of solute are needed to make 37.5 mL of 0.750 M KI solution? Round to three significant digits.

Marysya12 [62]

4.648 gm of solute is needed to make 37.5 mL of 0.750 M KI solution.

Solution:

We will start with the Molarity

$\text { Molarity }=\text { Mole of solute } \div \text { liter of solution }$

Also we know 1000 ml = 1 L

Therefore 37.5 ml by 1000ml we obtained 0.0375L

Equation for solving mole of solute

$\text { Mole of solute }=\text { Molarity } \times \text { Liters of solution }$

Now, multiply 0.750M by 0.0375

Substitute the known values in the above equation we get

$0.750 \times 0.0375=0.0281$

Also we know that Molar mass of KI is 166 g/mol

So divide the molar mass value to get the no of grams.

$0.028 \times 166=4.648$

So 4.648 gm of Solute is required for make 37.5 mL of 0.750 M KI solution.

8 0

3 years ago

What is the molarity of a solution that is made by mixing 35.5 g of Ba(OH)2 in 325 ml of solution?

choli [55]

Answer:

$M=0.638M$

Explanation:

Hello!

In this case, since the molarity of a solution is calculated by diving the moles of solute by the volume of solution in liters, we first compute the moles of barium hydroxide in 35.5 g as shown below:

$n=35.5g Ba(OH)_2*\frac{1molBa(OH)_2}{171.34gBa(OH)_2}\\\\n=0.207mol$

Then, the liters of solution:

$V=325mL*\frac{1L}{1000mL} =0.325L$

Finally, the molarity turns out:

$M=\frac{0.207mol}{0.325L}\\\\M=0.638M$

Best regards!

5 0

2 years ago